# Among any $2n$ consecutive integers below $n^2+2n$ at least one has no prime divisor less than $n$?

What is your idea about my conjecture?

Consider a sequence of $2n$ consecutive natural numbers, all the terms less than $n^2 + 2 n$. Then there exists at least one number in the sequence which is not divisible by any prime less than or equal to $n$.

• Bertrand's only talks about the interval (n,2n) – some one Dec 5 '15 at 15:05

The conjecture is equivalent to the following:

Among any $2n$ consecutive integers below $n^2+2n$, there is at least one prime greater than $n$.

Indeed: if your conjecture is true and none of the numbers in the sequence would be prime, then at least one number is the product of at least two primes $>n$, so is at least $(n+1)^2$, contradiction.

In particular it says:

There is at least one prime among $n^2,\ldots,n^2+2n-1$

which (because $n^2$ and $n^2+2n$ are never prime) is the same as

There is at least one prime among $n^2+1,\ldots,n^2+2n$

which is Legendre's conjecture and is unsolved.

• I know Legendre's conjecture.I'm looking for a counterexample to my conjecture – some one Dec 6 '15 at 9:45